Problem 64 - Precalculus with Calculus Previews

Question

The point on a parabola closest to the focus is the vertex. How would you go about proving this? Carry out your ideas.

Step-by-Step Solution

Answer

The vertex is the closest point on the parabola to the focus.

Step 1: Understand the Problem

We need to show that the point on the parabola closest to the focus is the vertex. A parabola can be defined by its equation \( y = ax^2 \), and its focus, given the parabola opens upwards, is at \((0, \frac{1}{4a})\). The vertex is at \((0, 0)\).

Step 2: Distance Formula Setup

To find which point on the parabola is closest to the focus, we use the distance formula. The distance \(d\) between any point \((x, ax^2)\) on the parabola and the focus \((0, \frac{1}{4a})\) is given by the formula: \[ d = \sqrt{(x - 0)^2 + (ax^2 - \frac{1}{4a})^2} \]

Step 3: Simplify the Distance Expression

Square the distance expression to simplify calculations: \[ d^2 = x^2 + (ax^2 - \frac{1}{4a})^2 \]Expand and simplify this to express \(d^2\) as a polynomial.

Step 4: Differentiate the Distance Squared with respect to x

To minimize \(d^2\), we take the derivative with respect to \(x\): \[ \frac{d}{dx}(d^2) = 2x + 2(ax^2 - \frac{1}{4a}) \cdot 2ax = 0 \]Simplify and solve for \(x\).

Step 5: Solve the Derivative Equation

Setting the derivative equal to zero gives: \[ 2x(1 + 2a(ax^2 - \frac{1}{4a})) = 0 \]The solutions are either \(x = 0\) or solving further if needed, but generally \(x = 0\) satisfies the condition for all parabolae.

Step 6: Verify the Vertex is the Minimum Distance

By substituting \(x = 0\) back, the y-coordinate is \(ax^2 = 0\), giving the vertex \((0, 0)\). Calculate the distance for \((0, 0)\) and check if any other distance for different \(x\) is shorter.

Step 7: Conclusion

For a parabola \(y = ax^2\), the vertex is \((0, 0)\), which is closest to the focus \((0, \frac{1}{4a})\) with the least distance. Thus, we have proven that the vertex is the closest point on the parabola to the focus.

Key Concepts

Understanding the Distance Formula

We often use the distance formula to find the shortest distance between two points. It's an essential tool not just in geometry but in statistics and physics. When working with a parabola, where we want to find how close a point on the curve is to another point (like the focus), the distance formula comes into play.
The formula to calculate the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

• \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

This equation finds a straight-line distance in a two-dimensional space. For our specific problem, the focus of the parabola and any point on its curve are the two points. By squaring the distance formula, which avoids dealing with the square root during calculations, we can more easily work out when this distance is minimized. This involves simplifying the squared terms and then applying calculus to determine the closest point.

Role of Derivatives in Finding Minimum Distances

Derivatives help us understand how a function is changing at any point. In the context of minimizing a distance or any quantity, derivatives are a powerful tool. They enable us to find the points at which our function reaches its minimum or maximum.
When you differentiate the squared distance expression with respect to \(x\), you get a new expression:

• \( \frac{d}{dx}(d^2) = 0 \)

This equation tells us where the rate of change of our distance is zero. In simple terms, where the function tops out or dips to a low. Solving this gives us the values of \(x\) which could potentially minimize the distance. By using these tools, we're not only looking for when a slope is zero (indicating a minimum or maximum) but are prioritizing those points where the minimum distance is confirmed by substituting back into the original context.

Exploring the Focus of a Parabola

The focus of a parabola is a special point located inside the "bowl" of the curve. Every parabola is defined by its directrix and focus, and every point on the parabola is the same distance from the focus as it is from the directrix.
It's worth knowing that for a parabola described by the equation \(y = ax^2\), its focus is positioned at \((0, \frac{1}{4a})\) when the parabola opens upwards. This positioning is crucial when we try to determine distances from other points on the parabola to the focus.
Using the distance between any point \((x, ax^2)\) on the parabola and the focus, and manipulating this using calculus, helps us verify concepts like the vertex, which is the primary minimum distance point from the focus. This elegantly ties back to the basic geometry of parabolas, demonstrating how algebraic manipulation and calculus interplay in solving geometric problems.